Question: $ E = \left[\begin{array}{rrr}1 & 4 & 5 \\ 3 & -2 & 0\end{array}\right]$ $ C = \left[\begin{array}{rr}4 & 5 \\ 0 & -2 \\ -2 & 4\end{array}\right]$ What is $ E C$ ?
Explanation: Because $ E$ has dimensions $(2\times3)$ and $ C$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ E C = \left[\begin{array}{rrr}{1} & {4} & {5} \\ {3} & {-2} & {0}\end{array}\right] \left[\begin{array}{rr}{4} & \color{#DF0030}{5} \\ {0} & \color{#DF0030}{-2} \\ {-2} & \color{#DF0030}{4}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ E$ , with the corresponding elements in column $j$ of the second matrix, $ C$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ E$ with the first element in ${\text{column }1}$ of $ C$ , then multiply the second element in ${\text{row }1}$ of $ E$ with the second element in ${\text{column }1}$ of $ C$ , and so on. Add the products together. $ \left[\begin{array}{rr}{1}\cdot{4}+{4}\cdot{0}+{5}\cdot{-2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ E$ with the corresponding elements in ${\text{column }1}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{4}+{4}\cdot{0}+{5}\cdot{-2} & ? \\ {3}\cdot{4}+{-2}\cdot{0}+{0}\cdot{-2} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ E$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{4}+{4}\cdot{0}+{5}\cdot{-2} & {1}\cdot\color{#DF0030}{5}+{4}\cdot\color{#DF0030}{-2}+{5}\cdot\color{#DF0030}{4} \\ {3}\cdot{4}+{-2}\cdot{0}+{0}\cdot{-2} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{1}\cdot{4}+{4}\cdot{0}+{5}\cdot{-2} & {1}\cdot\color{#DF0030}{5}+{4}\cdot\color{#DF0030}{-2}+{5}\cdot\color{#DF0030}{4} \\ {3}\cdot{4}+{-2}\cdot{0}+{0}\cdot{-2} & {3}\cdot\color{#DF0030}{5}+{-2}\cdot\color{#DF0030}{-2}+{0}\cdot\color{#DF0030}{4}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-6 & 17 \\ 12 & 19\end{array}\right] $